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A336362
Number of iterations of map k -> k*sigma(p^e)/p^e needed to reach a power of 2, where p is the smallest odd prime factor of k and e is its exponent, when starting from k = n. a(n) = -1 if number of the form 2^k is never reached.
5
0, 0, 1, 0, 2, 1, 1, 0, 3, 2, 2, 1, 2, 1, 3, 0, 4, 3, 3, 2, 2, 2, 2, 1, 2, 2, 3, 1, 4, 3, 1, 0, 3, 4, 3, 3, 4, 3, 3, 2, 3, 2, 3, 2, 5, 2, 2, 1, 5, 2, 5, 2, 4, 3, 4, 1, 4, 4, 4, 3, 2, 1, 4, 0, 4, 3, 5, 4, 3, 3, 4, 3, 5, 4, 3, 3, 3, 3, 3, 2, 6, 3, 3, 2, 6, 3, 5, 2, 6, 5, 3, 2, 2, 2, 5, 1, 6, 5, 5, 2, 6, 5, 3, 2, 4
OFFSET
1,5
COMMENTS
Informally: starting from k=n, keep on replacing p^e, the maximal power of the smallest odd prime factor in k, with (1 + p + p^2 + ... + p^e), until a power of 2 is reached. Sequence counts the steps needed.
LINKS
FORMULA
If A209229(n) = 1 [when n is a power of 2], a(n) = 0, otherwise a(n) = 1 + a(sigma(A336650(n))*(n/A336650(n))).
a(n) = a(2n) = a(A000265(n)).
EXAMPLE
For n = 15 = 3*5, we obtain the following path, when starting from k = n, and when we always replace the maximal power of the lowest odd prime factor, p^e of k with sigma(p^e) = (1 + p + p^2 + ... + p^e) in the prime factorization k: 3^1 * 5^1 -> (1+3)*5 = 20 = 2^2 * 5 -> 4 * (1+5) = 24 = 2^3 * 3^1 -> 2^3 * 2^2 = 2^5, thus it took three iterations to reach a power of two, and a(15) = 3.
PROG
(PARI)
A336650(n) = if(!bitand(n, n-1), 1, my(f=factor(n>>valuation(n, 2))); f[1, 1]^f[1, 2]);
A336362(n) = if(!bitand(n, n-1), 0, my(pe=A336650(n)); 1+A336362((n/pe)*sigma(pe)));
CROSSREFS
KEYWORD
nonn
AUTHOR
Antti Karttunen, Jul 30 2020
STATUS
approved